A solid cylinder of mass M and of radius R is fixed on a frictionless axle over a well. A rope negligible mass is wrapped around the cylinder. A bucket of uniform mass m is suspended from it. The linear acceleration bucket will:

Options

A. `(mg)/(M+m)`
B. `(2mg)/(M+m)`
C. `(2mg)/(M+2m)`
D. `(2mg)/(2M+m)`

Answer

To find the linear acceleration of the bucket suspended from the solid cylinder, we can follow these steps:

### Step 1: Analyze the Forces Acting on the Bucket
The forces acting on the bucket of mass \( m \) are:
- The gravitational force acting downwards: \( F_g = mg \)
- The tension \( T \) in the rope acting upwards.

Using Newton's second law, we can write the equation for the bucket:
\[ mg - T = ma \quad \text{(1)} \]
where \( a \) is the linear acceleration of the bucket.

### Step 2: Analyze the Torque on the Cylinder
The tension \( T \) in the rope creates a torque \( \tau \) on the cylinder. The torque can be expressed as:
\[ \tau = T \cdot R \]
where \( R \) is the radius of the cylinder.

### Step 3: Relate Torque to Angular Acceleration
The torque is also related to the angular acceleration \( \alpha \) of the cylinder by the equation:
\[ \tau = I \cdot \alpha \]
where \( I \) is the moment of inertia of the cylinder. For a solid cylinder, the moment of inertia is given by:
\[ I = \frac{1}{2} M R^2 \]
Thus, we can write:
\[ T \cdot R = \frac{1}{2} M R^2 \cdot \alpha \quad \text{(2)} \]

### Step 4: Relate Linear Acceleration to Angular Acceleration
The linear acceleration \( a \) of the bucket is related to the angular acceleration \( \alpha \) of the cylinder by the equation:
\[ a = R \cdot \alpha \]
From this, we can express \( \alpha \) as:
\[ \alpha = \frac{a}{R} \quad \text{(3)} \]

### Step 5: Substitute \( \alpha \) into the Torque Equation
Substituting equation (3) into equation (2):
\[ T \cdot R = \frac{1}{2} M R^2 \cdot \frac{a}{R} \]
This simplifies to:
\[ T = \frac{1}{2} M a \quad \text{(4)} \]

### Step 6: Substitute \( T \) into the Bucket's Equation
Now, substitute equation (4) into equation (1):
\[ mg - \frac{1}{2} M a = ma \]
Rearranging gives:
\[ mg = ma + \frac{1}{2} M a \]
Factoring out \( a \):
\[ mg = a \left( m + \frac{1}{2} M \right) \]

### Step 7: Solve for the Linear Acceleration \( a \)
Now, we can solve for \( a \):
\[ a = \frac{mg}{m + \frac{1}{2} M} \]

### Final Result
The linear acceleration of the bucket is:
\[ a = \frac{mg}{m + \frac{1}{2} M} \]

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